Can you find the shaded area | A Very Nice Geometry Problem

[@ناصريناصر-س4ب]

The area is not related to the radius of the small quarter circle. Let us make the radius zero, and from this the shaded area becomes equal to π*(6²)/4-π*(3²/2)=9π/2.

[@soli9mana-soli4953]

And the opposite extreme case is that one in which the radius of the semicircle is zero, then the smaller quarter circle is equal to the larger one, the radius =3 and area is again 9/2pi

[@ناصريناصر-س4ب]

Yes, I was satisfied with one case and left the other cases to other commentators so that I would not be selfish.​@@soli9mana-soli4953

[@ناصريناصر-س4ب]

​@@soli9mana-soli4953Yes, I was satisfied with one case and left the other cases to the rest of the commentators so that I would not be selfish.

[@うっちゃん-e8e]

OP₌OQ₌rとするとOB₌6₋r、AP₌6₋2r　1/4円AOBの面積：π(6₋r)^2/4　1/4円POQの面積：πr^2/4
1/2円APの面積：π(3₋r)^2より求める面積は(π/4){(6₋r)^2₋2(3₋r)^2₊r^2}＝9π/2

[@marioalb9726]

A = ½(¼πd²) = ⅛π6²
A = 4,5π cm² ( Solved √ )
or 2nd method :
A = ¼πR² - ½πr² = ¼πR² - ½π(½R)²
A = ¼πR² - ⅛πR² = ⅛πR² = ⅛π6²
A = 4,5π cm² ( Solved √)

[@五十嵐特許事務所]

Since the length of the AP is not given, the area of ​​the shaded region is considered to be constant regardless of the length of the AP. If AP = 0, the area to be found is the area of ​​a semicircle with a diameter of 6. Area = 9π/2.

[@oscarcastaneda5310]

I grew the smaller quarter circle to the be size of the larger quarter circle so that there's no semicircle at all and ... the required area is 9π/2 since doing this makes the figure a semicircle of radius 3.

[@СтасМ-ъ8б]

Пусть OQ = x= OP, AP=y, OB=R.
Очевидно x+R=6, x+y=R.
Сложим эти 2 уравнения:
2x +y=6, или x+y/2=3. Возведем обе части уравнения в квадрат:
x² + xy +y²/4=9. (i)
Уравнение x +y = R тоже возведем в квадрат:
x² +2xy+y²= R², или
xy= R²/2- x²/2- y²/2 .
Подставляем xy в уравнение (i) , получим:
R²/2+ x² - (y/2)²=9.
Умножаем обе части на π/2,
πR²/4 +πx²/4- π(y/2)²/2= 9π/2.
Очевидно, что в левой части площадь искомой фигуры, и она равна 9π/2.

[@kateknowles8055]

Thank you again for a morphing sort of puzzle. ( It felt like jelly out of its mould.)
.......... Taking the mid point of BQ as centre and drawing a lower semicircle of radius 3, for artistic reasons (not math this time)!
Now, being more diligent, let b = AO =BO q = OP = OQ and AP = 2.a
Quadrant OPQ has area ( pi/4) q.q Add the large quadrant and subtract the semicircle.
Required area (pi/4) (b.b + q.q - 2.a.a )
b + q = 6 (given) 2a + q = b 2a = b-q and therefore (b-q)(b+q) = b.b - q.q = 6.2.a = 12 a
Now the question does not specify the ratios of these three circles so the position of P above O will not affect the answer?
If q=0 Q, P, O are imposed on each other , then 2a =6 = b ( pi/4 ) (b.b + q.q - 2.a.a ) = (pi/4) ( 36 +0 - 6.3) = (pi/4) (36 -18)
If q =1 then b=5 and a =2 because 2a +1 = 5 (pi/4) (b.b +q.q - 2.a.a) = ( pi/4) (25 +1 -8) =( pi/2) (18/2)
For q =2 b=4 a=1 as in the diagram but these measurement are not given there (pi/2)(16+4-2)/2
And for q=3 b= 3 and a=0
And for any value of q between 0 and 3 this area is 9pi/2

[@zawatsky]

Рассуждаем логически. Полукруг АР, его площадь πr²/2. Четверть круга OPQ, его площадь π(2r)²/4=π4r²/4=πr² - две площади вырезанного сегмента или полный круг от АР. 2r составляет треть от BQ, отсюда r=6/6=1. Рисуем более удобную фигуру - обычную четверть круга + полукруг радиусом r. OB=6-2=4. Считаем. S(AOB)=π4²/4=4π. S(AP)=π1²/2=π/2. Итого общая площадь 4½π.

[@giuseppemalaguti435]

Ablue=(1/4)πR^2+(1/4)πr^2-(1/2)π((R-r)/2)^2=(1/4)π(R^2+r^2)-(1/8)π(R-r)^2=(1/8)π(2R^2+2r^2-R^2-r^2+2rR)=(1/8)π(R^2+r^2+2rR)=(1/8)π6^2=(9/2)π

[@RealQinnMalloryu4]

(6)^2 ➖ 36 90°APOB/36=2.18APOB 2.2^9 1.2^3^2 1.2^3^1 2^3 (APOB ➖ 3APOB+2).

[@RAG981]

Nice.